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Ksju [112]
3 years ago
13

Tridil is infusing at 15 ml/hr on an infusion pump. The drug is mixed 50 mg in 500 ml D5W. How many MCG/minute is the patient re

ceiving?
Physics
1 answer:
olga nikolaevna [1]3 years ago
8 0

Answer:

patient receiving drug 25 MCG/minute

Explanation:

given data

infusing = 15 ml/hr

drug = 50 mg

D5W = 500 ml

to find out

How many MCG/minute

solution

we know infusing rate is 15 ml/hr = 0.25 ml/min

so 0.25 ml drug content = 50 /500 × 0.25

0.25 ml drug content = 0.025 mg

so here

rate of drug will be 0.025 mg

rate of drug = 0.025 mg = 25 ×10^{-6} gm/min

rate of drug = 25 MCG/minute

so patient receiving drug 25 MCG/minute

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Three metal spheres are placed as shown in the image. Sphere A weighs 5 kg, B weighs 8 kg, and C weighs 3 kg.
Anni [7]

Answer:

B

Explanation:

7 0
2 years ago
A 3.35 kg object initially moving in the positive x direction with a velocity of 4.90 m s collides with and sticks to a 1.88 kg
ahrayia [7]

Answer:

The final components of velocity of the composite object is 3.33 m/s.

Explanation:

Given;

mass of the first object, m₁ = 3.35 kg

initial velocity of the first object, u₁ = 4.90 m/s in positive x-direction

mass of the second object, m₂ = 1.88 kg

initial velocity of the second object, u₂ = 3.12 m/s in negative y-direction

initial momentum of the first object, P₁ = 3.35 x 4.9 = 16.415 kgm/s

initial momentum of the second object, P₂ = 1.88 x 3.12 = 5.8656 kgm/s

The resultant velocity of the two objects is given by;

R² = 16.415² + 5.8656²

R² = 303.858

R = √303.858

R = 17.432 kgm/s

Apply the principle of conservation of linear momentum for inelastic collision;

total initial momentum before = total final momentum after collision

P₁(x) + P₂(y) = Pf

R = Pf

R = v(m₁ + m₂)

17.432 = v(m₁ + m₂)

where;

v is the final components of velocity of the composite object

v = \frac{17.432}{m_1 + m_2} \\\\v = \frac{17.432}{3.35+1.88} \\\\v = 3.33 \ m/s

Therefore, the final components of velocity of the composite object is 3.33 m/s.

8 0
3 years ago
Two particles, with charges of 20.0 nC and -20.0 nC, are fixed at points with coordinates <0, 4.00 cm> and <0, -4.00 cm
Dmitry [639]

Answer:

Explanation:

Potential energy of the system of charges

=  9 x 10⁹ x [  q₁q₂ / r₁₂ +  q₂q₃ / r₂₃ +  q₁q₃ / r₁₃ ]

here  r₁₂ ,  r₂₃ , r₁₃ are distance between 1 st and 2 nd charge , 2 nd and 3 rd charge and fist and third charge.

r₁₂ = 8 cm , r₂₃ = 4 cm , r₁₃ = 4 cm.

q₁ = 20 x 10⁻⁹ C , q₂ = - 20 x 10⁻⁹ C , q₃ = 10 x 10⁻⁹ C

Potential energy  =  9 x 10⁹ x [ - 400 x 10⁻¹⁸ / .08  +  -200x10⁻¹⁸ / .04 +  200 x 10¹⁸ / .04 ]

= 9 x 10⁹  x  - 400 x 10⁻¹⁸ / .08

= 45 x 10⁻⁶ J .

b)

Potential at the point of fourth charge due to three charges of 20 nC , - 20 nC and 10 nC at the centre

9 x 10⁹ [ 20 x 10⁻⁹ / .05 + - 20 x 10⁻⁹ / .05 + 10 x 10⁻⁹ / .03 ]

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= 3000 x 40 x 10⁻⁹ = 12 x 10⁻⁵ J .

kinetic energy at infinity = 12 x 10⁻⁵ J

1/2 m v² = 12 x 10⁻⁵ J

.5 x 2 x 10⁻¹³ x v² = 12 x 10⁻⁵

v² = 12 x 10⁸

v = 3.46 x 10⁴ m/s

= 9 x 10⁹

5 0
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Which color is in the middle of the light spectrum?
Elanso [62]
Well, with the light spectrum there technically is no middle color. Both green and yellow meet up in the middle at 560 nm (wavelength interval) and 540 THz (frequency interval).
3 0
3 years ago
6. Show that the weight of an object on the moon is 1/6 its weight on earth.​
mojhsa [17]

Taking ratio of W & w. ≈ 6 . w = 1/6 W. Therefore , Weight of an object on the moon is 1/6 of its weight on the earth.

5 0
3 years ago
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